Editing Gaussian elimination (section)

== Pseudocode ==
<!--{{Unreferenced section|date=March 2018}}-->
As explained above, Gaussian elimination transforms a given {{math|''m'' × ''n''}} matrix {{mvar|A}} into a matrix in [[row-echelon form]].

In the following [[pseudocode]], <code>A[i, j]</code> denotes the entry of the matrix {{mvar|A}} in row {{mvar|i}} and column {{mvar|j}} with the indices starting from&nbsp;1. The transformation is performed ''in place'', meaning that the original matrix is lost for being eventually replaced by its row-echelon form.

 h := 1 /* ''Initialization of the pivot row'' */
 k := 1 /* ''Initialization of the pivot column'' */
 
 '''while''' h ≤ m '''and''' k ≤ n
     /* ''Find the k-th pivot:'' */
     i_max := [[argmax]] (i = h ... m, abs(A[i, k]))
     '''if''' A[i_max, k] = 0
         /* ''No pivot in this column, pass to next column'' */
         k := k + 1
     '''else'''
         '''swap rows'''(h, i_max)
         /* ''Do for all rows below pivot:'' */
         '''for''' i = h + 1 ... m:
             f := A[i, k] / A[h, k]
             /* ''Fill with zeros the lower part of pivot column:'' */
             A[i, k] := 0
             /* ''Do for all remaining elements in current row:'' */
             '''for''' j = k + 1 ... n:
                 A[i, j] := A[i, j] - A[h, j] * f
         /* ''Increase pivot row and column'' */
         h := h + 1
         k := k + 1

This algorithm differs slightly from the one discussed earlier, by choosing a pivot with largest [[absolute value]]. Such a ''partial pivoting'' may be required if, at the pivot place, the entry of the matrix is zero. In any case, choosing the largest possible absolute value of the pivot improves the [[numerical stability]] of the algorithm, when floating point is used for representing numbers.<ref>{{Cite journal |last1=Kurgalin |first1=Sergei |last2=Borzunov |first2=Sergei |date=2021 |title=Algebra and Geometry with Python |url=https://doi.org/10.1007/978-3-030-61541-3 |journal=SpringerLink |location=Cham |language=en |doi=10.1007/978-3-030-61541-3|isbn=978-3-030-61540-6 }}</ref>

Upon completion of this procedure the matrix will be in row echelon form and the corresponding system may be solved by back substitution.